1. **Problem Statement:** We have a joint density function for random variables $Y_1$ and $Y_2$ given by: $$f(y_1, y_2) = \begin{cases} \frac{1}{2}, & 0 \leq y_1 \leq y_2 \leq 2 \\ 0, & \text{otherwise} \end{cases}$$ We want to find: (a) The conditional density of $Y_1$ given $Y_2 = y_2$. (b) The probability that less than $\frac{1}{2}$ gallon is sold given $Y_2 = 1.5$. 2. **Step (a): Find the conditional density $f_{Y_1|Y_2}(y_1|y_2)$** - The conditional density is defined as: $$f_{Y_1|Y_2}(y_1|y_2) = \frac{f(y_1, y_2)}{f_{Y_2}(y_2)}$$ - First, find the marginal density of $Y_2$: $$f_{Y_2}(y_2) = \int_0^{y_2} f(y_1, y_2) \, dy_1 = \int_0^{y_2} \frac{1}{2} \, dy_1 = \frac{y_2}{2}$$ - This is valid for $0 \leq y_2 \leq 2$. - Therefore, $$f_{Y_1|Y_2}(y_1|y_2) = \frac{\frac{1}{2}}{\frac{y_2}{2}} = \frac{1}{y_2}$$ - The support for $Y_1$ given $Y_2 = y_2$ is $0 \leq y_1 \leq y_2$. 3. **Step (b): Calculate $P(Y_1 < \frac{1}{2} | Y_2 = 1.5)$** - Using the conditional density: $$P\left(Y_1 < \frac{1}{2} \middle| Y_2 = 1.5\right) = \int_0^{1/2} f_{Y_1|Y_2}(y_1|1.5) \, dy_1 = \int_0^{1/2} \frac{1}{1.5} \, dy_1 = \frac{1}{1.5} \times \frac{1}{2} = \frac{1}{3}$$ **Final answers:** - (a) $$f_{Y_1|Y_2}(y_1|y_2) = \frac{1}{y_2}, \quad 0 \leq y_1 \leq y_2$$ - (b) $$P\left(Y_1 < \frac{1}{2} | Y_2 = 1.5\right) = \frac{1}{3}$$